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magicjohn2278 Special user Isle of Man UK 544 Posts |
Well put lockedroomguy!
I agree! |
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leonard Regular user North Carolina 148 Posts |
Lockedroomguy,
The order matters in the sense that it helps us keep the different outcomes separate. If I flip a coin twice in a row, the four possible outcomes are: HH, HT, TH, TT. If I flip two identical coins, the possible outcomes are the same. However, the second and third outcomes appear identical. Without some way to distinguish between these outcomes (i.e. order, different coins, etc.), the tendency is to view HT and TH as the same outcome. Nevertheless, the "mixed" outcome (HT or TH) remains twice as likely as the other two possibilities (HH or TT). Maybe this will help. I have two dogs, one of them is a male. What is the probability that the other one is also a male? Answer: 33%. I have two dogs, the older one is a male. What is the probability that the other one is also a male? Answer: 50%. I have two dogs, the taller one is a male. What is the probability that the other one is also a male? Answer: 50%. I have two dogs, the black one (assuming only one is black) is a male. What is the probability that the other one is also a male? Answer: 50%. You get the idea. regards, leonard |
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magicjohn2278 Special user Isle of Man UK 544 Posts |
Quote:
On 2006-03-04 12:06, leonard wrote: So how do you work that out!? |
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magicjohn2278 Special user Isle of Man UK 544 Posts |
Quote:
On 2006-03-04 12:06, leonard wrote: This is wrong, the "mixed" outcome is just as likely as the "two the same" outcome. If you consider HH and TT to be "two possibilities", then by your own argument, you can't consider HT and TH to be the same (the "mixed" outcome)! |
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magicjohn2278 Special user Isle of Man UK 544 Posts |
I'm loving this!!!
Quote:
On 2006-03-04 12:06, leonard wrote: I have two dogs, the male one is a male. What is the probability that the other one is also a male? Answer: ??????%. |
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leonard Regular user North Carolina 148 Posts |
Magicjohn2278,
I have two dogs. (Possible outcomes - MM, MF, FM, FF.) One of them is male. (Possible outcomes - MM, MF, FM.) (FF is eliminated.) What is the probability that the other one is also a male? One outcome (MM) out of three (MM, MF, FM) equals 33%. regards, leonard P.S. I'm off to finish Nir's problem in another thread. |
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magicjohn2278 Special user Isle of Man UK 544 Posts |
Leonard, if you have two dogs, There are only THREE possible outcomes - Either they are both male (Mm), they are both female (Ff) or they are one of each (Mf)
If we know one dog is male we disregard the (Ff) so it is a 50 / 50 chance that the other dog is a male. QED and Regards John |
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leonard Regular user North Carolina 148 Posts |
John,
I have a related question. I have six-hundred fair coins (50/50 chance of heads/tails) and three buckets. I flip two coins at a time. If the result is HH, both coins go into bucket #1. If the result is TT, both go into bucket #2. If the result is mixed, both go into bucket #3. How many coins are in each bucket after all three-hundred pair have been flipped? regards, leonard |
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lockedroomguy New user New York, NY 65 Posts |
Leonard,
Thanks for the answer. I think you may be emphasizing my question, though. It is common to use the initial letter system when working on coin probability puzzles, and it is important there to differentiate between, say, HTT and THT. What I'm not clear on is why, in this puzzle, the order matters. |
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LobowolfXXX Inner circle La Famiglia 1196 Posts |
Quote:
On 2006-03-04 11:44, lockedroomguy wrote: The difference in labeling "MF" differently than "FM" is for clarity, ultimately, but it may not look that way. Essentially, it's to emphasize that the "one of each" possibility is twice as likely as the "both male" possibility (or the "both female" possibility). Let's say the dogs are a terrier and a beagle. It's true that if you just look at the numbers, there are only "three possibilities": Both male, both female, or 1 of each. However, if you look at all possible outcomes, there are "four possibilities": Both male, both female, male terrier & female beagle, and female terrier & male beagle. Labeling "MF" distinctly from "FM" serves two purposes: it distinguishes between two actually different situations (older dog male & younger female against vice versa, or however else you distinguish the two), AND it emphasizes that the "1 and 1" possibility is, in fact, twice as likely as either of the same sex possibilities.
"Torture doesn't work" lol
Guess they forgot to tell Bill Buckley. "...as we reason and love, we are able to hope. And hope enables us to resist those things that would enslave us." |
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LobowolfXXX Inner circle La Famiglia 1196 Posts |
The key to the puzzle posted, and puzzles like this, is this: How was the information obtained? It has practical application in, for instance, competitive bridge (where a variation of the theme is called the principle of restricted choice) as well as the old "Let's Make a Deal" game show, where knowledgeable contestants could have doubled their odds of winning the "good" prize. Some information, by the nature of its revelation, leaves the a priori odds untouched; other information signficantly changes them.
"Torture doesn't work" lol
Guess they forgot to tell Bill Buckley. "...as we reason and love, we are able to hope. And hope enables us to resist those things that would enslave us." |
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dlhoyt Regular user 176 Posts |
Quote:
On 2006-03-04 20:42, leonard wrote: To answer leonard's question, 150 coins are expected to be in bucket #1 (HH); 300 coins in bucket #2 (HT); 150 coins in bucket #3 (TT). If this doesn't make intuitive sense consider this minor modification to leonard's problem: Start with 300 fair pennies and 300 fair nickles. Flip them two at a time, but flip a penny with a nickle each time you do. If both coins are heads put them in bucket 1; if both are tails they go into bucket 3; if you get 1 head and 1 tail, IRRESPECTIVE OF WHICH IS WHICH, put them bucket 2. This should make it easier to see why the HT combination occurs twice as often as the HH or TT combinations. (You can get one head and one tail two ways: the penny is H and the nickle is T or the penny is T and the nickle is H.) Nothing changes if the coins are all the same, just substitute first coin for the penny and second coin for the nickle. Half the time you'll get one head and one tail and 1/4 the time you'll get two tails and 1/4 the time you'll get two heads. Hope this helped. Dale |
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magicjohn2278 Special user Isle of Man UK 544 Posts |
Leonard and all, I'm beginning to see your point of view...
HOWEVER, I think that from the original question, it is safe to say, that the guy with the dogs isn't going to shop at the place where they dish them out at random, or determine what you are going to get by the toss of a coin! Oh no! This guy is particularly interested in the sex of his pets! (We won't ask why.) We know this because it forms the majority of the question in the first place. He gets his pets from the place that offers you a choice of sex! "Two dogs please". "What sort would you like?". "What are the options?", he asks. "Well you can have two females, two males, or one of each." "Which is most popular?". "None of them, we seem to sell them all in equal amounts." |
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Sergey Smirnov New user Belgium 70 Posts |
If we take all the people who have 2 dogs, then about 1/4 of them will have 2 male dogs, 1/4 2 female dogs, and 1/2 of them will have dogs of both sexes. If people having FF are eliminated, then the ratio becomes 1/3 vs 2/3, hence 1/3 probability. Hope this helps.
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magicjohn2278 Special user Isle of Man UK 544 Posts |
Thanks Sergey, I DO see the logic..... (I can't say that I AGREE with it!)
Incidentally, I went to a shop yesterday to buy one cat and one parrot, I had to come home empty-handed, they had sold out of cats and parrots, they only had one parrot and one cat left!? |
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Sergey Smirnov New user Belgium 70 Posts |
So what is your take on how many there are 2 male dog pairs, 2 female dog pairs and pairs with dogs of both sexes?
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Sergey Smirnov New user Belgium 70 Posts |
So I suppose your answer will be 1/3 (MM), 1/3 (FF), and 1/3 (both sexes). Only in this case the odds of the other dog being male are 50%. Now let me show why it is wrong.
Take all the pairs of dogs owned by people. How many pairs of dogs are there in which the older dog is male? 50%, right? Of those 50% how many pairs are there in which the younger dog is female? Also 50%. 50% of 50% is 25%. E.i. there are 25% of pairs in which the older dog is male and the younger dog is female. By the same reasoning there are also 25% of pairs in which the younger dog is male and the older dog is female? And those pairs are different from the first ones. 25%+25% is 50%. Therefore every second pair of dogs consists of dogs of different sexes. |
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LobowolfXXX Inner circle La Famiglia 1196 Posts |
Depends on the parameters of the question...in the case John presented, where the pet store guy says, "Well, there are three options: both male, both female, and one of each, all equally popular," then one COULD randomize the pairs and, for instance, roll a 6-sided die: 1-2 = both male; 3-4 = both female; 5-6 = one of each. In this case, the odds of any of the three choices would be equally likely, and in the long run, the pet store would go through as many male as female dogs, so they wouldn't care. If you randomize the PAIRS, all are equally likely. If you randomize the DOGS, then the mixed pair is twice as likely as either of the same-sex pairs.
"Torture doesn't work" lol
Guess they forgot to tell Bill Buckley. "...as we reason and love, we are able to hope. And hope enables us to resist those things that would enslave us." |
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Sergey Smirnov New user Belgium 70 Posts |
Yes, it's true. In the original problem we have the second situation, though.
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magicjohn2278 Special user Isle of Man UK 544 Posts |
Sergey, I quite happy to accept that the distribution of pairs of dogs might be as you describe, but it doesn't help your problem!
Possibilities: 1 MM 2 MF 3 FM 4 FF There are 8 dogs in the table above. We know that line 4 (FF) doesn't exist. But if we simply cross it out, we have eliminated two female dogs from the table. Which seems a little unfair! Lets assume that your friend didn't buy them as a pair (Only options are in equal distributions of 2 male, 2 female, or one of each - which you don't accept!) He bought one then the other. And in the possibilities above assume the MF or FM illustrates the order that he bought them in. So we agree that line 4 can't exist in the problem, but it isn't as simple as just deleting it. The reason line 4 doesn't exist, is because one of his dogs is male. - But we don't know whether it was the first dog that he bought, or the second. (If he bought them as a pair then you have to accept the original argument!) So, when he says that one of his dogs is a male, is he referring to the first or the second? We don't know, but it has to be one of them! If we assume that it is the first dog he is referring to, then the only remaining possibilities are: 1 MM 2 MF 3 4 If it is the second dog he is referring to, then the only remaining possibilities are: 1 MM 2 3 FM 4 In both cases giving a 50/50 chance that the other dog is a male!! (It might be easier to look at it another way, we are eliminating a female, either the first, or the second. (But not both)) |
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